Answer
$\frac{{{{\tan }^3}\theta }}{3} + \frac{{{{\tan }^5}\theta }}{5} + C$
Work Step by Step
$$\eqalign{
&\text{Let }I= \int {{{\tan }^2}\theta } {\sec ^4}\theta d\theta \cr
& {\text{Split }}{\sec ^4}\theta {\text{ as }}{\sec ^2}\theta {\sec ^2}\theta \cr
& I = \int {{{\tan }^2}\theta } {\sec ^2}\theta {\sec ^2}\theta d\theta \cr
& {\text{Rewrite }}{\sec ^2}\theta {\text{ using }}{\sec ^2}\theta = 1 + {\tan ^2}\theta \cr
& I = \int {{{\tan }^2}\theta } \left( {1 + {{\tan }^2}\theta } \right){\sec ^2}\theta d\theta \cr
& = \int {\left( {{{\tan }^2}\theta + {{\tan }^4}\theta } \right)} {\sec ^2}\theta d\theta \cr
& {\text{Let }}u = \tan \theta ,{\text{ }}du = {\sec ^2}\theta d\theta \cr
& {\text{Substituting}} \cr
& I = \int {\left( {{u^2} + {u^4}} \right)du} \cr
& {\text{Use the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& I= \frac{{{u^3}}}{3} + \frac{{{u^5}}}{5} + C \cr
& {\text{Write in terms of }}\theta ,{\text{ substitute }}\tan \theta {\text{ for }}u \cr
& I = \frac{{{{\tan }^3}\theta }}{3} + \frac{{{{\tan }^5}\theta }}{5} + C \cr} $$
